The smallest natural number n, such that the coefficient of x in the expansion of `(x^(2) + (1)/(x^(3)))^(n)` is `.^(n)C_(23)`, is

Options

A. 35
B. 23
C. 58
D. 38

Explanation

To find the smallest natural number \( n \) such that the coefficient of \( x \) in the expansion of \( (x^2 + \frac{1}{x^3})^n \) is \( \binom{n}{23} \), we can follow these steps:

### Step 1: Write the general term of the expansion
The general term \( T_k \) in the expansion of \( (x^2 + \frac{1}{x^3})^n \) is given by:
\[ T_k = \binom{n}{k} (x^2)^{n-k} \left(\frac{1}{x^3}\right)^k \]
This simplifies to:
\[ T_k = \binom{n}{k} x^{2(n-k)} x^{-3k} = \binom{n}{k} x^{2n - 2k - 3k} = \binom{n}{k} x^{2n - 5k} \]

### Step 2: Find the condition for the coefficient of \( x \)
We want the power of \( x \) to be 1:
\[ 2n - 5k = 1 \]
Rearranging gives:
\[ 5k = 2n - 1 \quad \Rightarrow \quad k = \frac{2n - 1}{5} \]

### Step 3: Set \( k \) as an integer
Since \( k \) must be a non-negative integer, \( 2n - 1 \) must be divisible by 5. Therefore, we can write:
\[ 2n - 1 \equiv 0 \mod 5 \quad \Rightarrow \quad 2n \equiv 1 \mod 5 \]

### Step 4: Solve for \( n \)
To solve \( 2n \equiv 1 \mod 5 \), we can find the multiplicative inverse of 2 modulo 5. The inverse of 2 modulo 5 is 3, since:
\[ 2 \cdot 3 \equiv 6 \equiv 1 \mod 5 \]
Multiplying both sides of \( 2n \equiv 1 \) by 3 gives:
\[ n \equiv 3 \mod 5 \]
Thus, \( n \) can be expressed as:
\[ n = 5k + 3 \quad \text{for some integer } k \]

### Step 5: Find the smallest natural number \( n \)
The smallest natural number \( n \) occurs when \( k = 0 \):
\[ n = 3 \]

### Step 6: Check if \( n = 3 \) satisfies the condition
We need to check if \( k = \frac{2(3) - 1}{5} = \frac{6 - 1}{5} = 1 \) is valid. Thus, we have:
\[ k = 1 \]
Now, we check:
\[ 2n - 5k = 2(3) - 5(1) = 6 - 5 = 1 \]
This is valid.

### Conclusion
The smallest natural number \( n \) such that the coefficient of \( x \) in the expansion of \( (x^2 + \frac{1}{x^3})^n \) is \( \binom{n}{23} \) is:
\[ \boxed{3} \]